Lesson 3.5
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School
Louisiana State University *
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Course
PRECALC
Subject
Mathematics
Date
Nov 24, 2024
Type
docx
Pages
18
Uploaded by jarjiiiii
Lesson 3.5
Answer:
An exponential growth model has the form y
=
ae
bx
and an exponential decay model has the form y
=
ae
−
bx
Answer:
A logarithmic model has the form y
=
a
+
b
ln
x
or y
=
a
+
b
log
x
Answer:
In probability and statistics, Gaussian models commonly represent populations that are normally distributed
.
Answer:
A logistic growth model has the form y
=
a
1
+
be
−
rx
Answer:
Time to double: 6.60 years
Amount after 10 years: $2143
Answer:
Annual % rate: 11%
Time to double: 6.3 years
Answer:
Initial Investment: P ≈
$1,122.47
Annual % Rate: 5.7762%
Answers:
(a)
n
=
1
t
=
ln
2000
1000
(
1
)
ln
(
1
+
0.065
1
)
t
=
ln2
ln
(
1.065
)
t
=
11
years
(b)
n
=
12
t
=
ln
2000
1000
(
12
)
ln
(
1
+
0.065
12
)
t
=
10.69
years
(c)
n
=
365
t
=
ln
2000
1000
(
365
)
ln
(
1
+
0.065
365
)
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t
=
ln 2
(
365
)
ln
(
1
+
0.065
365
)
t
=
10.66
years
(d)
When compounded continuously
2000
=
1000
e
0.065
t
Dividing both sides by 1000
2
=
ln
e
0.065
t
Taking natural logarithm of both sides
ln 2
=
ln
e
0.065
t
Using the inverse property ln
e
x
=
x
ln 2
=
0.065
t
t
=
ln2
0.065
t
=
10.66
years
Answer:
a
=
6.5
g
y
=
1
2
6.5
y
=
3.25
g
Therefore,
3.25
=
6.5
e
−
b
(
5715
)
Dividing both sides by 6.5
3.25
6.5
=
e
−
b
(
5715
)
1
2
=
e
−
b
(
5715
)
Taking natural logarithm of both the sides
ln
1
2
=
ln
e
−
b
(
5715
)
Using the inverse property ln
e
x
=
x
ln
1
2
=−
5715
b
Dividing both sides by (-5715)
b
=
−
1
5715
ln
1
2
Using the value of b found above, a
=
6.5
, and t
=
1000
years, the amount left is
y
=
6.5
e
−
(
ln
1
2
5715
)
1000
y
=
6.5
e
(
ln
1
2
5715
)
1000
y
=
5.76
g
Answer:
Using the above model and put the values to get the initial amount,
A
2
=
Ae
−
24100
k
Solve the above equation for k
,
A
2
=
Ae
−
24100
k
1
2
=
e
−
24100
k
Take logarithm on both sides,
log
1
2
=
log
e
−
24100
k
log
1
2
=
¿
−
24100
k
¿
Dividing by 24100 on both sides,
log
1
2
24100
=
24100
k
24100
2.8
×
10
−
5
=
k
Amount left after 1000 years is 0.4 gm. Hence,
A
(
1000
)
=
0.4
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t
=
1000
k
=
2.8
×
10
−
5
Using the above parameters and exponential decay model,
0.4
=
Ae
−
1000
×
2.8
×
10
−
5
¿
A e
−
0.028
Solve the equation for A,
0.4
=
Ae
−
0.028
0.4
=
A×
0.972
0.4114
=
A
Hence the initial amount is 0.4114
gm.
Answer:
(a)
P
=
347
and t
=
6
347
=
319.2
e
6
k
e
6
k
=
347
319.2
¿
1.087
ln
e
6
k
=
ln1.087
6
k
=
0.0834
k
=
0.0139
As k
is positive, the population is increasing
(b)
Now to find the population in the year 2015, let t = 15
P
=
319.2
e
kt
¿
319.2
e
0.0139
×
15
¿
319.2
×
1.2318
¿
393.1998
So the population of Tallahassee in the year 2015 will be
≈
393,200
Now to find the population in the year 2020, let t = 20
P
=
319.2
e
kt
¿
319.2
e
0.0139
×
20
¿
319.2
×
1.3205
¿
421.499
So the population of Tallahassee in the year 2020 will be
≈
421,499
(c)
410
=
319.2
e
0.0139
t
e
0.0139
t
=
410
319.2
¿
1.2845
ln
e
0.0139
t
=
ln1.2845
0.0139
t
=
0.2504
t
=
18.014
So, the population will reach 410,000 when t
is approximately 18. That is, in the year 2018
Answer:
The initial population of the bacteria given is
a
=
250
The population after 1 hour is
t
=
1
therefore,
y
=
250
e
b
(
1
)
The population after 10 hours is double the population at 1 hour.
t
=
10
Therefore,
2
(
250
e
b
)
=
250
e
b
(
10
)
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2
=
e
9
b
ln 2
=
ln
e
9
b
Using the inverse property ln
e
x
=
x
ln 2
=
9
b
b
=
ln 2
9
b
=
0.077
Now, the number of bacteria after 6 hours is
y
=
250
e
0.077
(
6
)
y
=
396.81
y ≈
397
Answers:
(a)
After 20 days on the job, a new employee produces 19 units
Therefore substitute t
=
20
and N
=
19
in the model,
N
=
30
(
1
−
e
kt
)
19
=
30
(
1
−
e
k
(
20
)
)
19
30
=
1
−
e
204
e
204
=
1
−
19
30
¿
11
30
20
k
=
ln
(
11
30
)
k
=
1
20
(−
1.0033
)
¿
−
0.501
Hence, the learning curve of the employee is N
=
30
(
1
−
e
−
0.0501
t
)
(b)
Now employee produces 25 units per day.
Therefore substitute N
=
25
in N
=
30
(
1
−
e
−
0.0501
t
)
25
=
30
(
1
−
e
−
0.0501
t
)
25
30
=
1
−
e
−
0.0501
t
e
−
0.0501
t
=
1
−
25
30
e
−
0.0501
t
=
ln
(
5
30
)
t
=
1
−
0.0501
(−
1.7917
)
¿
35.76
≈
36
Hence, 36 days should pass before this employee is producing 25 units per day.
Answer:
Modern and ancient charcoal contains same amount of 14 C. A piece of charcoal contains only 15% as much radioactive carbon as a piece of modern charcoal.
Hence, A
(
t
)
=(
0.15
)
A
0
Therefore, calculate time t
as below,
(
0.15
)
A
0
=
A
0
¿
(
0.15
)
=
¿
log
(
0.15
)
=
t
5715
log
(
1
2
)
log
(
0.15
)
=
t
5715
log
(
0.5
)
t
5715
=
log
(
0.15
)
log
(
0.5
)
t
=(
5715
)
−
0.8239
−
0.3010
¿
(
5715
)(
2.7372
)
≈
15642
Hence, the tree burnt 15642 years ago to make the ancient charcoal.
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Answers:
(a)
In the year 1998, t = 0,
P
=
2632
1
+
0.083
e
0
¿
2632
1.083
≈
2,430,286
people
In the year 2005, t = 5,
P
=
2632
1
+
0.083
e
0.50
×
5
¿
2632
1.10657
≈
2,378,512
people
In the year 2010, t = 10,
P
=
2632
1
+
0.083
e
0.50
×
10
¿
2632
1.10657
≈
2,315,182
people
(b)
2500
2450
2200
2000
0
2
4
6
8
10
12
14
16
18
20
(c)
From the graph it is clear that the curve and the green line intersect at
(
17.2
,
2,200
)
.
So the year in which the population will reach 2.2 million is 2017.
(d)
To check the validity of part (c) algebraically, let P = 2200 and solve for t.
2200
=
2632
1
+
0.083
e
0.050
t
1
+
0.083
e
0.050
t
=
2632
2200
So, 0.083
e
0.050
t
=
0.1963636
e
0.050
t
=
0.1963636
0.083
≈
2.365826
0.050
t ≈
ln
(
¿
2.365826
)
¿
≈
0.8611276
t ≈
17.22
So, the year in which the population will reach 2.2 million is 2017.
Answer:
(a)
We have to find the magnitude, R,
given I
=
199,500,000.
Now,
R
=
log
199,500,000
1
Therefore,
R
=
8.3
(b)
We have to find the magnitude, R,
given I
=
48,275
,
000.
Now,
R
=
log
48,275,000
1
Therefore,
R
=
7.7
(c)
We have to find the magnitude, R,
given I
=
17,000
.
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Now,
R
=
log
17
,
000
1
Therefore,
R
=
4.2
Answer:
This is a logarithmic model since, x
=
0
,
log
x
=
0
Answer:
This is a logistic growth model since; it becomes constant after a while.
Answer:
This is an exponential decay model.
Answer:
This graph is a linear model.
Answer:
This graph is an exponential model.
Answer:
This is a quadratic model, as the value of x is positive or negative, the graph will take the values in positive or negative quadrant for x.
Answer:
This is a Gaussian model.
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